text:
Can Instrument Approaches be carried out when RVR is not given for the full runway length? (Approach Ban)
Bristol Airport (EGGD) has the following NOTAM in place:

INSTRUMENTED RVR RWY 27 STOPEND OUT OF SERVICE

Would aircraft be allowed to continue approaches into Bristol? If so, would the touchdown and mid-point RVR values suffice to decide whether the approach is legal (see Approach Ban below).
For example, the listed RVR minima for the RNP Approach runway 27 is 1400m (LNAV) - if the touchdown point and mid point RVR values were above this, could someone legally proceed with the published approach.
The UK AIP describes the Approach Ban as follows.

AD 4.9  Approach Ban - All Aircraft
4.9.3 An aircraft may commence an instrument approach regardless of the reported RVR/Visibility but the approach shall not be continued
below 1,000 FT above the aerodrome if the relevant RVR/Visibility for
that runway is at the time less than the specified minimum for
landing.
...
4.9.6 The touchdown zone Runway Visual Range (RVR) shall always be controlling.

To perform an ILS CAT I approach, at least the touchdown zone RVR sensor must be operational. For CAT II, you need the touchdown zone and midfield sensors, and for CAT III you need all three. As per your NOTAM, if only the stopend RVR sensor is out of service, it should be possible to perform CAT I and CAT II approaches (assuming those are available for that runway in the first place).
There are procedures which allow aircrews to perform a CAT I ILS based on meteorological visiblity alone, using a special conversion table to estimate RVR values based on visiblity, but I believe those are company specific.
RVR can also be measured manually, but I am not sure this is actually allowed anymore. It would require someone to physically go to the end of the runway, and then count the number of runway lights visible from the end. Given there is a fixed distance between runway lights, you could then multiply the number of visible lights with the distance between them to get the RVR.
Not sure what the rules are for RNP approaches, but I would imagine they are similar to those regarding ILS.

In the U.S., under Part 121, you can fly an approach Cat 1 with all RVR transmissometers inop as long as the prevailing visibility is at/above the minimum required (typically 1/2 SM if everything else like approach lights is working). The first half sentence of your 2nd paragraph is spot-on; the rest, not as much.

@RalphJ Probably depends on the country.  Feel free to downvote and/or provide an alternative answer

Many thanks. @expeditedescent You can also ascertain RVR from a visibility conversion: “Where the RVR is not available, RVR values may be derived by converting the reported visibility” (CAT.OP.MPA.305 ). This derivation is achieved through the conversion table: EASA AMC10 CAT.OP.MPA.110 Table 8 Conversion of reported MET VIS to CMV.


text:
Is this an accurate method to showing how the correct Decision Altitude/Height for an approach has been determined?
Here's my method of summarising how the decision altitude/height for an approach was determined:
DH/A =
The highest base altitude/height from the list below +
Altimeter Pressure error +
Altimeter Temperature Error +
Company Operational Procedures Additions

Base altitude/height:

OCA/H
System Minima e.g. ILS 200ft, VOR/DME 250
OCA/H per aircraft category
(Otherwise) Published approach DA/H
Minimum A/H to which the approach aid can be used
Minimum A/H in the Aircraft Flight Manual

Instrument Rated pilots will need an understanding of how the DA was determined, they may be examined on it in theory tests and skills proficiency tests - though whilst actually flying the approach they will use the plate DA.
Is this summary correct, or how would you alter it?

The correct method is to hire a mechanic to fix all that error in your altimeter. Then set it to the local airport pressure setting. I don't know where you are flying but around here we don't have unverified approach navaids or published terminal procedures without a listed DA/MDA. Without a published procedure it is either a standard VFR approach or your company has developed its own approach which will be verified and have a DA or MDA. Why or how would an aircraft manual have procedure altitudes?

Generic system minima is used by the engineer designing an approach, it really has no common use for pilots, all systems will have a published approach with listed minima. IMC flight below MOCA requires a published approach. A visual approach has no DA/MDA.

@MichaelHall flying

@MaxPower Instrument rated pilots need to understand how the DA on a plate is determined (not just unknowingly accept it without comprehending how it is determined). This formula was my attempt to remember.

@ob318, while I agree it is useful to understand why certain things are the way they are, when you are flying an approach to mins you don’t need to “determine” the DA, it has already been determined, and published.  If you are designing an approach that’s a different story.  In that case the published is irrelevant because you haven’t determined what it should be yet.  Right?  You are mixing things in a confusing way and I am going to vote to close this until you can clarify the question.

@MichaelHall of course I don’t need to determine the DA whilst on the approach. But CPL IR rated and ATPL pilots are expected to have the theoretical knowledge to know how the DA was determined. This is the premise of the question. For both theory and practical examinations, CPL and ATPL require the ability to demonstrate this working knowledge. I am just posting this question to check whether the my knoweldge is correct, as organsied into a formula.

@ob318, again I am not disagreeing that background knowledge is good, but your question is still a confusing blend...

@MaxPower many aircraft types will have altimeters which differ from ISA conditions, particularly those without ADCs. Furthermore temperature errors around zero and sub zero temperatures are well accepted, and listed in the AIP of each state.

@MichaelHall how would you un-confuse it? I have added more detail to the question.

@ob318, sorry, I think I made that last comment before fully reading and understanding your edit.  Much better.  The way it read before it seemed as if you were espousing an “AVEF”-like approach to pilot recomputation of a more correct DH to fly than what is published.  I removed my downvote and close vote.  I think you are correct, but I don’t know enough about approach design to confirm or expand on your understanding, so I will leave it at that...

I directly submitted an approach design question to the FAA to clarify a change in terminology in one of the primary procedure design regulations. They did answer my question but also stated that they don't include definitions because those design criteria are only intended for the specific government department that creates procedures. In other words, pilots are not required to know this level of detail.

@MaxPower EASA pilots do, sadly.

This is USA information. The summary: determine the obstacle clearance surfaces, add the required obstacle clearance buffers to those surfaces, adjust DA for glide slope angle and aircraft speed if not already above the minimum due to obstacle clearance. The procedures are all calculated in true altitude, not indicated altitude. Aircraft that fly IFR in the USA are required to have adjustable precision altimeters based on the ISA atmosphere and barometric altimeter based approaches must have a local weather station that provides the altimeter setting for that airport. The difference in DA based on temperature is not considered significant, however there are low temperature limits placed on baro-vnav approaches due to effects on glide path [not DA].
A typical DA in the USA is determined without altimeter temperature adjustments because a DA is only used with positive vertical course guidance, a three degree slope is the same at any temperature. There may be be a fixed decision-altitude adjustment calculated when a remote weather source must be used, based on potential difference in pressure. There may also be a low temperature limit in some cases for procedures that allow uncompensated baro-vnav systems because this equipment can change the glide path angle.
While there were some old manual calculation standards that could be simplified into the answer you are looking for, the modern method is calculated by computers using rather complex algorithms with multiple intersecting surfaces and measurement-uncertainty adjustments as well as large databases of terrain, obstacles, and local winds. FAA order 8260.3D contains most of the approach design structure, it is 509 pages, but even with all of that in several places it simply references the use of software, an example quote "Increase ROC values by the amount specified by the software." See 8260.3D page 198, 202-203,(cat1 final obstacle surfaces and DA) and appendix C(precipitous terrain adjustments) for some of the formulas.
This is completed with a detailed in flight inspection of the procedure using specialized aircraft to carefully map the navigation signals and controlling obstacles. The FAA also performs this service for many other countries and performs regular flight checks of existing procedures, nav-aids, and airways on a regular schedule. Some basic procedure/airway design requirements can be altered if a flight inspection determines the deviation is safe in that specific case.
As for the core components, they check obstacle clearance of the final approach segment, and recently they also compute the vertical guidance surface.(the part of final between DA and threshold. Note the 34:1 clearance symbol added to new FAA plates, it is related to VGS. Jeppeson omits this piece.) If this requires an increased glide slope angle to meet required clearances the DA may be raised for the faster category aircraft. Then the missed segment is calculated and checked for obstacles, if needed the DA is raised again so that the missed segment has required clearances. The missed segment includes an inertia allowance below DA for aircraft that fail to react until they are at DA,(this is a buffer, not intended for normal use.) this inertia is also why higher speed aircraft are given a higher DA. The difference in height between the navigation equipment in the cockpit, or antenna placement, and the wheels of the aircraft-class expected at that airport is considered for the area between DA and threshold.(A 747 will have its wheels about 30 feet lower than a cessna 172 while following the same glide path.)

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text:
ILS Designated Operational Coverage (DOC)
During a standard Instrument Landing System (ILS) approach, I typically arm the VOR/LOC (Boeing 737) to capture the localiser1 once on an intercept course. Upon clearance for the approach, I then arm APP mode to capture the glideslope.
According to my company's Standard Operating Procedures (SOPs), APP mode can only be armed once cleared for the approach and within the Designated Operational Coverage (DOC) of the ILS.
My theory textbooks define ILS coverage as extending "18 NM within 10° of either side of the course centreline and 10 NM within 35° of the course centreline."
Is this the typical DOC that my company SOPs refer to?
Typically, I arm the approach mode approximately 10-12 NM from landing. However, a fellow pilot I flew with recently insisted that APP mode should be no armed further than 10 NM due to DOC. Can someone clarify the typical DOC requirements for ILS approaches
For example, see this NOTAM which describes the DOC into Bergamo as being 25nm/10000ft for IBRM ILS RWY 28:

A4884/24 NOTAMN
Q)LIMM/QIDXX/I/NBO/A/000/999/4540N00942E005
A) LIME
B) 2406121049 C) PERM
E) DME 'IBRM' ASSOCIATED WITH ILS RWY 28 DESIGNATED
OPERATIONAL COVERAGE: 25 NM/10000FT REF AIP AD 2 .19 'RADIO NAVIGATION
AND LANDING AIDS' COLUMN 7 'DESIGNATED OPERATIONAL COVERAGE
LIMITATIONS'

1: American editors, please resist the urge to correct the spelling. I am a British English user and localiser is an acceptable spelling. Thanks

Wouldn't a question about company SOPs be best addressed by someone at your company?

@FreeMan While it's true that company SOPs reference the Designated Operational Coverage (DOC), the DOC itself is a standard defined by aviation authorities and not by the company. My question is focused on understanding the typical DOC requirements for ILS approaches, as these are not unique to my company but rather are applicable to aviation operations in general. Clarifying the DOC helps ensure adherence to correct procedures and enhances overall safety. Hence, I'm seeking a broader understanding of DOC in the context of ILS approaches rather than specific company guidelines.

Might be best to edit that explanation into the question itself.

@FreeMan I do explicitly say that my company SOPs "refer" to DOC. I talk about SOPs to bring context to the area, but DOC is referred to as something outside of the SOPs, and not defined by the company.

The Designated Operational Coverage area is to protect the ILS frequency from interference by other ILS installations. The DOC is defined in ICAO EUR Doc 11 (Frequency Management Manual).
A DOC is defined for each transmitting component of the ILS; the localiser, the glidepath and the DME.
For arming the APP mode to capture the glide path, the DOC of the glide path is relevant here.
The Glide Path is protected in an area within 8 degrees from the localiser centreline out to a distance of 10NM.

I believe that supervisory authorities responsible for managing the spectrum and/or verifying the correct functioning of the navaids can define DOCs that are not in line with the standard above. In such case, the actual DOC needs to be published in the AIP.
For example, the German AIP publishes the 'operational coverage' or 'usable' range and sector of navaids associated to an airport in tables of section AIP.AD.2.

thanks, though you do see NOTAMs (now added to bottom of the question) suggesting a much higher DOC than 10nm.

@tedioustortoise I suspect that the authorities in charge of frequency protection are allowed to define DOCs that deviate from the standard DOC. I think DOCs are published in the AIP, but I don't know exactly in which section. Note that the NOTAM in your post specifically mentions the DME associated to the ILS, not the ILS itself. The 25NM DOC is thus for the DME, not for the localiser or glidepath; they may have different DOCs.

I just checked the German AIP for Frankfurt (EDDF). In the Aerodrome (AD) section there is a table listing all the navaids associated with the airport and in column 8: Remarks the operational coverage or usable range/sector is described. For the ILSes (both LOC and GP)  they all look non-standard.

@tedioustortoise This answer refers to the glide path DOC only, your NOTAM refers to the DME DOC. The localizer DOC is different again (see my answer).

The ILS coverage is specified in ICAO Annex 10. What your textbook says corresponds to the reduced localizer coverage already, normal coverage is larger:

3.1.3.3.1 The localizer shall provide signals sufficient to allow satisfactory operation of a typical aircraft installation
within the localizer and glide path coverage sectors. The localizer coverage sector shall extend from the centre of the
localizer antenna system to distances of:

46.3 km (25 NM) within plus or minus 10 degrees from the front course line;
31.5 km (17 NM) between 10 degrees and 35 degrees from the front course line;
18.5 km (10 NM) outside of plus or minus 35 degrees from the front course line if coverage is provided;

except that, where topographical features dictate or operational requirements permit, the limits may be reduced down to
33.3 km (18 NM) within the plus or minus 10-degree sector and 18.5 km (10 NM) within the remainder of the coverage when
alternative navigational means provide satisfactory coverage within the intermediate approach area.

That is for the localizer, the glide path coverage is smaller, maybe that is what your colleague was referring to:

3.1.5.3.1 The glide path equipment shall provide signals sufficient to allow satisfactory operation of a typical aircraft
installation in sectors of 8 degrees in azimuth on each side of the centre line of the ILS glide path, to a distance of at least
18.5 km (10 NM) up to 1.75 θ and down to 0.45 θ above the horizontal or to such lower angle, down to 0.30 θ, as required to
safeguard the promulgated glide path intercept procedure.

Thanks. Just to clarify, is this DOC? Operational coverage, and not theoretical coverage?

@tedioustortoise ICAO Annex 10 doesn't actually use the term "Designated Operational Coverage" at all, it just defines Coverage. Other ICAO sources (example, see page 4) then use the term DOC for the coverage defined in Annex 10.

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text:
Why is fuel quantity expressed in flight hours
Going through a few aircraft documentaries online, I've noticed the thing that they usually express the fuel quantity in flight hours.
Ex - A 747 burns 4 litres of fuel every second. So with a ton of fuel, it'll stay in air for roughly 3-4 minutes.
Why are flight hours a necessary term while consider fuel quantity? I mean shouldn't they be worried about how much distance they can cover in an emergency to reach an airport, instead of how long they can stay in the air?
Seems a question related to endurance but couldn't find any proper explanation

It’s primarily due to wind.
Aircraft move through the air at a certain speed, which makes it tempting to give range in miles (or kilometers), but the air itself is moving as well, and often at a large enough fraction of the aircraft’s airspeed that it materially affects range across the ground.
For instance, say you have a Cessna 172 with four hours of fuel (plus reserves) and a cruise speed of 110kt. With a 30kt tailwind (140kt ground speed), your range is 560nm. But going the opposite direction with a 30kt headwind (80kt ground speed), your range is only 320nm—just over half as far!
To make it even more complicated, the winds will have different directions and speeds at different altitudes and locations, and all of that changes over time too, which is one of the reasons pilots (or dispatchers) have to carefully plan each flight’s route and fuel load, even for a “fixed” daily route between the same two airports.

In other vehicles, like automobiles, measuring fuel quantity in miles (or kilometers) makes more sense because there is usually a fairly direct correlation between fuel burn and mileage. This is not the case with aircraft. The distance that a plane can travel on a single tank of gas will vary widely, based on atmospheric conditions.
Headwinds may not affect cars very much, but a 40 knot headwind can just about cut a small plane's cruise speed by half. Prop planes don't work as efficiently at high altitudes, jets don't work very efficiently at low altitudes. It just makes more sense for a pilot to know how long the engines will keep running - then they can calculate the available distance based on the current conditions.

Is it possible for you to add an example of how this is used in real life? Say an aircraft flying Eastward at 350 knots with 10 knots of headwinds, how/ what will I get out of it exactly? Pretty sure that's not all the required information but I hope you got my problem

I'd think that "measuring fuel quantity makes more sense" in any vehicle. Something missing there?

@Salmonbeing00 A 10 knot wind is fairly irrelevant, but the high altitude Jet Stream winds have been measured at up to 275 knots. On one occasion a flight from LA to London landed almost an hour ahead of schedule because of a 200 knot tail wind, and the schedule would already include the effect of an "average" 100mph wind adding about an hour to a 6 hour flight from east to west compared with still air, and reducing it by an hour flying from west to east.

@Salmonbeing00, what specific help do you need ?  The math is really quite simple, just subtract headwind or add tailwind.  In the specific example you offered, at the end of an hour instead of traveling 350 nautical miles, with a 10 knot headwind you would have traveled 340. That’s what you will get out of it.  Speed is just distance over time.  Is this what you are asking?

The unit of measure used for fuel is entirely dependent on the context:

When discussing endurance or talking to ATC you refer to fuel in flight time.
When calculating weight and balance you refer to it by weight (kilograms or pounds).
When computing cost or talking about tank capacity you refer to it by volume (liters or gallons).

As far as air traffic control is concerned, fuel endurance is the measure which air traffic control uses to initiate search and rescue if the aircraft becomes overdue. It is the most practical method for an aircraft with which communication has been lost.

Well, there's a simple way to think about it.
A plane with no fuel will do only one thing, fall or just not move (the one leads to the other anyway)
So fuel keeps us up there and without it, we're pretty much doomed.
Given the cruise, climb and descent speeds and the known distance we calculate the time it will take to get to places. If we need to redirect to another destination for some reason we need to know how much longer we can stay in the air which means fuel in liters or gallons isn't useful.
So we rather calculate how long the engine can keep us airborne using fuel flow values in different flight sections and the total fuel to determine this.

-1: planes do not "fall" when they run out of fuel. Besides being obvious knowledge for anyone who flies, noticeable examples are the Gimli Glider, US Airways 1549, and many others.

Gliding is falling, only very slowly and with a horizontal component ;) With the right thermals you can even fall upwards.

@Michael: no. When an aircraft is gliding, forces are equalized and there's no acceleration. Vertical speed is steady in calm air (with a decent pilot, at least). You wouldn't say you are "falling" when you walk down the stairs. Most thermals also require decent pilots ;)


text:
How should night time be calculated when it is not night during the entire flight?
I am B777 rated pilot and software developer. Nowadays I am working on an application about flights and this app has to calculate night time. Still I am trying to figure out how to calculate night time correctly for logging purposes. As you know night definition is easy,

Night means the time between the end of evening civil twilight and the
beginning of morning civil twilight, as published in the Air Almanac,
converted to local time.

But calculating a flight's night time is a little bit confusing.

Is this just related about being day or night at takeoff and landing airports?
Or whole the route should be considered if it is day or night at that time while flying?

I searched a lot but couldn't find any formula or detailed explanation about this. Just found some sites and apps can calculate night time; but interestingly every app gives different result; and still it is unclear for me that how they calculate night time.
Example:
Departure : KMIA
Destination : LTBA
Block Off Time : 03:03Z
Block On Time : 13:36Z
Block Time : 10:33

If you calculate this flight's night time via http://www.crewlogbook.com/utilities/night_calculator.php you will get:
Night Time : 01:47

I wonder how this calculation was made?

The second part of your question(s) is how to calculate the actual night time. I am afraid this is off-topic here. You might want to remove that part of the question. Just as a hint, calculating night time is a not so trivial astronomical calculation. There is no simple formula for it. There are a number of software libraries freely available, like this one for Javascript: https://github.com/mourner/suncalc . Similar libraries exist for many other languages.

About different results from different apps: all these calculations are numerical solutions of simplified models. There is no precise result. Some apps deliver a better approximation, some give a more coarse one. The authors of the libraries should document how accurate their results are. Differences of a few minutes may be considered acceptable for the sake of a fast computation.

I would be very surprised if no night flight time counts at all for the flight that starts early evening and finishes late morning on another day. For a long range aircraft like 777 this is probably not unusual.

@bogl my purpose of calculating night time is for logging. As you may know pilots should log their flights' night times to their logbooks. So calculating that particular flight's night time becoming important. Probably because my English the question was not clear, i was not asking how can i calculate night time for a place or city, I am trying to figure out how can I can calculate flight's night time. I hope it is more clear now

@h22 we have such routes that either we always fly towards sun in daylight or fleeing from sun in night around 12 hours. So you are correct sometimes I fly 13 hour and couldn't log any night time.

I'm not in the position to give a qualified answer. From my gut feeling, logging sunset and sunrise by observation should be accurate enough for this purpose.

@bogl thank you for your interest. It is interesting that there is no clear explanation or formula about this. I was expecting that FAA or EASA published an approved calculation method unfortunately they didn't.

I don't think they could even if they wanted to. It is complicated to calculate for a stationary location. Calculating for a moving object makes it even more computing intensive. There is no formula or explanation simple enough to publish, and understandable for pilots without sufficient math background. They would have to publish an algorithm, and you would need a software implementation to run it. I doubt that the purpose of night time logging justifies such an effort.

@bogl actually it is clear that there is calculation method; because there are apps or sites able to calculate your flight’s night time according to your flight data. I am trying to find how do they calculate?

Yes, of course, software is able to do it for you. This is an example of how it works: https://github.com/mourner/suncalc/blob/master/suncalc.js .

@bogl yes this library is amazing but I think we have still misunderstandings. This library can provide me sunset and sunrise hours for a coordinate, this is OK. The question is how should I use this information to correctly calculate night time for that particular flight. I am adding an example to the question.

@Pondlife thank you for suggestion; I already checked it before asking this question. Unfortunately that question look similar but it is different also the answers are not what I am looking for.

Bahadir: You would have to calculate sunset/sunrise for virtually every point along your track. Then compare each result with the time when you passed through that point. When the times matches the sunrise or sunset time, you found a transition point and record it. The duration of night along your track is the difference between the sunrise time on the track and the sunset time on the track (modulo 24h). This is for algorithms, not for people with pen and paper.

@bahadir arslan: While I'm not a commercial pilot, I suspect most do it the way I do: look out the window and see if it's dark, then ballpark the number.  If you're doing it algorithmically, it could get really complex, as you'd have to have detailed flight track info, and do interpolation for sun positions...

@bogl yes, this is what I am looking for. Ralph J's answer is also similar. So I am accepting it and thank you for your support. Now I will start writing code to calculate night flight time.

By the time that you're operating a 777, the amount of "night" time in the logbook is probably of little to no concern, so getting things exactly right to the minute may not be a great worry.
I'd simply grab a location every 15 minutes, either by GPS if your app has that, or else as (time elapsed / total time) * total great circle distance, and then for that location determine if you're between EECT & BMCT at that time. If so, log those minutes as night.
You could get fancier by interpolating the BMCT (for instance) between the last point you were in "night" and the first point you weren't. Then you have "the" time that you passed from night into day.
To implement that, you'll need to calculate the EECT & BMCT for an given point, plus find points along the great circle route. I'd expect that algorithms are available for both of those functions, but it's not something I've ever implemented.

This is not a cruicial for me. But I am developing an app so for that I have to do this. Your suggestion sounds logical. I wish I could found an offical explanation about night flight time.

Look out the window:  If it’s black outside, log night time. If it’s gray outside, log actual IMC. Everything else is day.

I have done many flights where the sun hovers around just at the horizon and will sometimes “rise”  and then “set” many times during the flight. Unless you have an exact route and an exact time during the entire flight, any App calculations will not be accurate for day or night.

@MikeSowsun you are right, I cannot make precise calculation without knowing actual route and timings. But this will be an approximate calculation. I will try my best.

@MichaelHall it will be an app so looking out the window is not an option. Thank you anyway

There is the ambient light sensor on the mobile device. It can be used for picking the information about the current level of light, telling this way if there is a night or a day. All you need then is to set the timer inside your program to add the duration of the last minute to the appropriate counters (total and also either day or night). Here is a complete tutorial about the luminosity detector on the Android device and here is another about the timer.
The problem remaining is to find the place where the device does not see much of artificial light from inside the aircraft.
Many mobile devices actually use "assisted GPS" (AGPS) that picks the satellite data from Internet rather than from the satellite itself. This kind positioning does not work well in flight mode.


text:
Why does airspeed increase with more throttle if RPM stays the same?
Given a constant speed propeller, the RPM will stay the same because of the governing system which changes the blade angle. But in cruise flight if you add more power and keep the RPM's the same, there is an increase in airspeed.
So when there is an increase in throttle/MP and more fuel/air mixture goes to the cylinders, does that extra combustive force attempt to spin the shaft faster and thus the blade angle increases and takes a "bigger bite of air"? Or is there some aspect of increased torque force pushing harder against the air?
I prefer the first explanation but I have heard some form of the second one being used, but in a more complete way. I honestly can't remember how that was described and I think I just misunderstood that explanation. What am I missing?

Re "Or is there some aspect of increased torque force pushing harder against the air?" -- why the word "or" in this sentence?  Aren't "attempt to spin the shaft faster"/ "takes a "bigger bite of air" and "increased torque force pushing harder against the air" fully compatible concepts?   Seems like this is essentially just two different ways of saying the same thing.

Could you perhaps improve the question by isolating one particular variable that you are trying to understand whether it stays constant, or doesn't, when the throttle position is changed?  LIke, I dunno, just brainstorming here, total momentum change of air flow as it moves through the prop disk?  Volume of air moved?  Average velocity change of air?  Or perhaps simply torque applied to prop shaft?  Don't want to rewrite your question for you, but seems like you have presented two sentences that are not clearly different, and are asking "which is true?"

Yes, your assumption as stated below is correct.  There's really no simpler way to explain it than that.

So when there is an increase in throttle/MP and more fuel/air mixture goes to the cylinders, does that extra combustive force attempt to spin the shaft faster and thus the blade angle increases and takes a "bigger bite of air"?


text:
Why does ATC assign a SID and then give different instructions after takeoff?
I have had many occasions at different airports flying a small single engine piston aircraft out of a Charlie or Delta and they will assign the SID, but give different instructions upon getting airborne. Is the SID assigned as a lost comms safety net? Or is there another reason why they assign the SID and then don’t actually have you fly it?
For example, flying out of KCHA, being assigned the Chattanooga Eight, and then being given completely different heading and altitude instructions open contacting departure. Same thing has happened at KAVL on the Asheville Seven.

Specific examples would help.

A SID, among other things, is a tool to ease ATC coordinations for departing traffic. For airports with established SIDs, it is common for the tower to be allowed to depart aircraft on the established SIDs, without explicitly coordinating (as in, picking up the phone) every single flight with approach control. Approach knows to expect traffic departing on the SIDs, and will keep those areas free of arriving or other aircraft. This saves a lot of time for both tower and approach, who would otherwise need to agree specifically on the departure instructions given to every single flight prior to departure.
Once you make it airborne, approach may well take you off the SID in order to expidite things for yourself and other traffic in the area. SIDs rarely give the most direct routing, as they are designed by default to be separated from other departure and arrival flows, military areas etc. If these restrictions are not relevant at the specific time you depart, a more direct routing may be possible. Or, maybe approach needs you off the SID because a following (faster moving) departure is coming up behind you.


text:
How to calculate wing upwash angle?
How can I calculate upwash angle(alpha u) at the wing? I notice at every position upwash angle is different, which angle is chosen as the reference upwash angle?

For a preliminary estimation of the upwash in front of the wing, the following plot¹ can be used:

Given the distance in front of the wing (x-axis) and the aspect ratio $A$, the plot returns $\frac{\partial \epsilon_u}{\partial \alpha}$ (y-axis), where $\epsilon_u$ is the upwash angle. The value $\frac{\partial \epsilon_u}{\partial \alpha}$ from the plot has to be multiplied by the wing's AoA to get $\epsilon_u$.

¹from: Daniel P. Raymer, Aircraft Design: A Conceptual Approach; in turn taken from: Hoak D. et al., USAF DATCOM.

Camber line at the bottom right is contour of wing?

Yes, from 0.25 till -0.75

Where can I find how to calculate upwash angles, lifting line, lifting surface theory?

Regarding upwash any book about general airplane design contains some method to estimate it. Regarding lifting line/surface theory any book about aerodynamics contains an explanation about it.


text:
What limits maximum Cl of wing with active jet blowing ("Blowing BLC")?
Maximum lift coefficient of wing (single-element or multiple-element) without suction or jet blowing are limited by boundary-layer separation(flow separation).
Wings that use jet blowing for boundary-layer-control(BLC) have no flow separation problems at upper surface, what then limits their maximum lift coefficient?
I can no longer find the source where I read it, but in short:
3D downwash effects is reason to limit max lift coefficent. Lift vector is tilted back and magnitude is reduced because vortex wake is tilted downward, so induced velocity has forward component that substract from freestream velocity.
Result is, at some point vertical component of lift stop increasing.
Can you clarify this?
Picture source

Picture source

I don't think that it will change the Cl. Cl is defined as lift per unit area AND per unit dynamic pressure; Cl is independent of dynamic pressure. From the figures, it seems that the jet is only increasing the dynamic pressure, which will increase lift, but not lift coefficient.

@AdityaSharma Cl is defined to freestream velocity not jet velocity.

Not true. Cl is strictly defined as lift per unit area and per dynamic pressure. Cl is practically constant for a particular aerofoil at a particular AoA. The jet increases the lift, but only by increasing the dynamic pressure; it has no effect on the Cl.

@sophit It is wing with "blowing BLC". Thanks, I will write source.

The paper from which you got the pictures does explain how such devices work plus it gives references to deepen the matter. Have a look there

@AdityaSharma "Cl is practically constant for a particular aerofoil at a particular AoA". Cl depends at least on the Reynolds number. And if compressibility is not negligible, it depends also on the Mach number. And if unsteady phenomena are important, it depends also on the reduced frequency. And for sure there are other particular dependencies for particular fields of application. In case of an aircraft Cl is normally defined in respect to the freestream velocity, unless otherwise specified.

What I mean with "practically constant Cl" is that the stall speed (dynamic pressure), for all practical purposes, does not depend on any speed related factors including Mach number or Reynolds number. In this statement, I assumed that the Mach number is low enough for compressibility effects to be neglected. And the reason that effects from Reynolds number can be neglected is that the operating speed range of any realistic aircraft is too low for Reynolds number to take an effect. Yes, compressibility effects cannot be neglected at higher Mach numbers, but for simplicity, lets ignore it.

@sophit My point is, the jet will increase the lift and the critical AoA without increasing the Cl; the critical AoA is increased due to the induced downwash from the jet. When this configuration is operated at its Cl-max, and the lift equation is observed, it is found that the dynamic pressure simply doesn't make up for the increase in lift. One may believe that the additional increase in lift is due to increased Cl, but in reality that lift has nothing to do with the aerofoil; that lift is actually the vertical component of thrust from the jet, and will be produced even without the aerofoil.

@AdityaSharma: the jet is normally blown out horizontally on the back of the wing, it has no vertical component, eventually only thrust. $C_l$ is defined in respect to the freestream velocity, full stop. Is this definition equivalent to cheating when analysis high-lift devices? Yes sure, but it gives a common base of comparison. Which q would you use to define $C_l$ in this case? The one of the freestream investing the forward part of the airfoil? The one of the jet? The one of the airflow at the trailing edge? Even the airfoil itself is different and couldn't be compared in the first place!

My first comment addressed the now-removed figure of a literal jet engine Jerry-rigged to the wing leading edge, and my following statements elaborated that particular case. What you are saying is absolutely true, if the jet is an INTEGRAL part of the wing (slotted flaps/slats), then it will increase the Cl. In fact, as you pointed out, the addition of the jets means that this aerofoil is entirely different from the one we started with. In the aforementioned case, the jet was NOT integral with the aerofoil, and only for that case, I claimed that the Cl (and Cl-max) won't change. Good day!

@AdityaSharma: ok, it wasn't clear that you were referring to a specific case and unfortunately modifications of the questions makes following someone's reasoning difficult

@sophit No issues! :)

The theoretical maximum lift of a cylinder is 4π. At that point the forward and rear stagnation points coalesce on the bottom of the cylinder. Practical solutions use rotating cylinders (Flettner rotor) but will not reach this theoretical limit. There is a way in potential flow theory to view the flow around an airfoil as that around a cylinder (conformal mapping), which helps to see what the theoretical limits for airfoils are.
This means your maximum lift coefficient will never be higher than 12.5. Since BLC will mostly help with shifting the rear stagnation point (and a bit with the forward, too), it will achieve quite a bit less.
Blowing will allow you to shift the rear stagnation point most of the way to the bottom of that cylinder (draw a circular control surface around your wing and view it as that theoretical cylinder). When the forward stagnation point is ahead at the front and the rear at the bottom, lift (from potential theory) will be 2π (approx. 6.3).
Maximizing the lift coefficient always means to avoid flow separation during recompression from as much suction as possible, and to do this recompression in the smallest length possible. Blowing helps with that, as does adding new boundary layers by introducing gaps between airfoil elements.
Adding a slat and a slotted flap to a regular airfoil will bring you at most to a c$_{l_{max}}$ of 2.8. Using multi-element flaps helps, as demonstrated by the airfoil of the Boeing 727. It had a triple slotted flap which could reach a two-dimensional c$_{l_{max}}$ of 4.2 with 60° deflection and rearward movement of the flaps which increased the effective wing area. Note that wing sweep and ailerons reduced the practical maximum lift coefficient of the full aircraft to 3.0 (see plot below, taken from A. M. O. Smith's excellent article High-Lift Aerodynamics).

Another practical limit is the maximum Mach number in the suction peak of all the airfoil elements. Once this reaches 1.58, no lift growth could be observed in experiments. This translates to a maximum for the product of Mach squared and pressure coefficient of -1.0. In other words: You need to fly very slowly in order to achieve high values of c$_{l_{max}}$.
The only practical design I know which reached that magic limit of 6.3 given as the c$_{l_{max}}$ of its wing airfoil is the Antonov An-70. Here this number is valid for the inner wing which is immersed in the prop wash of the mighty Aerosila SV-27 propellers, driven by the 13,240 hp Progress D-27 turboprop engines. By using the flight speed and clean wing chord for the reference parameters, this looks to me a bit like cheating, but so is your blowing.
If you look from afar such that your wing becomes a point, the oncoming flow will bump straight into the wing (actually, it first rises up and hits the wing with the upwash angle), and leave the wing with a downwash angle which is twice as large as the upwash angle. In effect, your wing will impart a momentum to the flow which is the product of the mass flow through a circle with a diameter equal to the span of the wing, the initial flow speed and the sine of the upwash angle.
Adding this momentum to the air causes a reaction force on the wing which we split in lift and drag. Since this force vector is tilted aft more when the downwash angle gets larger, drag is nonzero and eats away a bit of the lift. But the reaction force will not become infinitely large, and even if it would, it still would have a vertical component, which then also would become infinitely large. Only when the reaction force points straight back will lift be limited.
In reality the downward speed component added by the wing is at most $$v_z = \frac{S\cdot c_L\cdot\frac{v^2}{2}\cdot\rho}{\frac{b^2}{4}\cdot\pi\cdot\rho\cdot v} = \frac{2\cdot c_L\cdot v}{\pi\cdot AR}$$
with $AR = \frac{b^2}{S}$ the aspect ratio of the wing. If we now insert that maximum lift coefficient from potential flow theory, we get $$v_z=\frac{8}{AR}\cdot v$$
With the aspect ratio of a typical airliner the downwash angle now would be 90°, making the angle by which the reaction force is tilted backwards 45°. This means lift is 70% of what the reaction force is - drag eats away some of the lift but still has a rather small effect. Still, L/D is only 1 at this point, so compared to normal flight drag is punishingly large.

I think @user0422 wants to know the why of those increments (energisation of the boundary layer or whatever fancy term is more appropriate) and not the how much. I think it would also make sense to briefly explain why a rotating cylinder with its 4π is our point of reference.

@sophit I guess I got overwhelmed by the multitude of pictures.

@PeterKampf Can I add to my question what I read as  reason but I didn't quite understand it?

A lot indeed, but CFD (coloured fluid dynamics) is always nice to see :-)

@user0422 Please go ahead and edit the question, so all is in one place. Let me know when you are done.

@PeterKämpf I added.

@user0422 Added my two cent, too.

@PeterKämpf "Only when the reaction force points straight back will lift be limited." If reaction force points back, then lift is zero?   Why in my explanation induced velocity reduce  freestream velocity due to tilt vortex wake? Does wing so much bend the flow,so upwash/downwash behaves like  wall/blockage for freestream, so freestream velocity is reduced?

@user0422 I tried to use this to show how absurd the concept of lift limited by drag is. Sorry to have confused you. Please simply forget that paragraph.

@PeterKämpf Can you just tell me, why tilted vortex wake reduce freestream velocity?

@user0422 Who says it reduces velocity? The wake is passive, it is not acting. The wing does. And the wing only changes the direction of the air flowing over and under it, but does not reduce it outside of the boundary layer. The result is the wake.

@PeterKämpf Thanks for the explanation.


text:
Landing gear drag and weight calculations
I have read numerous books and researches, but I couldn’t find any formulae that actually work for how to calculate the following (for commercial aircrafts):
1- Drag coefficient for a retractable landing gear.
I tried the following equations:
CD,gear = WingLoading  * FlapDeflection * (( AircraftMass(TO,LD))^-0.215)
CD0,gear = SUM i=1 > n [CD,gear(AreaLG / WingArea)]
The results were ASTRONOMICAL
2- Weight of main and nose landing gears.
LG weights are calculated as a percentage of the MTOW, but different sources had different percentages (Normally between 3.8% and 6%). That doesn’t really seem to be the case; For example, a Boeing 747–400BCF’s MTOW is 410 tonnes. Based on my research, the wing-mounted gears (2) are 2952 kg EACH, body-mounted gears (2) are 2924 kg EACH, and the nose gear is 1432 kg. In total, main landing gears weigh 11752 kgs, which is, roughly, 2.9%. If the nose landing gear weight is added to the calculation that’d be 13184, which is 3.2% for ALL landing gears combined!
If anyone knows how to calculate the CD or weights, please help. Very much appreciated!
Cheers

How did you decide that the drag coefficient of the landing gear would vary based on Wing loading or flap deflection or aircraft mass?  Cd basically says how much force is exerted on the gear as the aircraft moves through the air (well, when multiplied by 1/2 rho V squared times the frontal area of the gear).  Similar to when you stick your hand out the window of a moving car and then either hold your hand palm down or palm forward you are changing the drag coefficient of your hand.  But that CD of your hand has nothing to do with how much the car weighs or whether the trunk is open.

I understand that. You are absolutely correct! That's why I said I couldn't find anything that actually works or makes sense. The formulae stated in the question were found in "Drag Force and Drag Coefficient: Chapter 3." It doesn't make sense, hence why the results were astronomical.

1. C$_D$
This data might be in the aeroplane design books in statistical form. A quick look in the one I have, Synthesis Of Subsonic Airplane Design by E. Torenbeek, has some data for slow, fixed wing aeroplanes in Appendix F.
Fluid-Dynamic Drag from Sighard Hoerner states the following on page 13-15:

Retractable Landing Gears (it seems) are often designed without much care for the aerodynamic shape in the extended position. The drag coefficient (based on wheel area) is in the order of $C_{D◻}$ == 0.6 or 0.8....for configurations with cylindrical struts.

The emphasis on Wheel Area in above citation is mine, the value would have to be converted to the wing area reference $C_D$.

A further source was found in my paper copy handout of Airplane Performance, VTH-D 14 by Prof. Wittenberg, for a Convair 240. The increase in $C_D$ is relatively constant over the $C_L$ range, about 0.02.
2. Weight
This one is in Torenbeek, page 280 gives an overview for several jet transports, relevant data 35 years ago but still giving a good general trend.

No idea which sources you are referencing, please be specific when referring to a reference.
You should check out Curry's book on landing gear design.
There is also an extensive report put together by one of Bill Mason's grad students.
Curry doesn't do a whole lot for drag, but it has a nice section on conceptual design level weights (Ch. 11) as well as more detailed treatment in the rest of the chapters.
His data shows that the total weight of all landing gear (main and nose) is about 3.5 to 4% of max landing weight (slightly less of max gross).  This is true across a wide swath of aircraft types.  He provides modifiers for longer/shorter gear and rough field requirements, etc.
Landing gear can be a huge contributor to an aircraft's drag.  Gear accounted for half the drag of a clean flying wing radio controlled aircraft I designed once.  The wheels and struts all likely have separated flow -- which is very draggy.
Use Hoerner and build up a drag estimate based on cylinders for the struts and the given data on wheels.  It isn't too complex


text:
Are there any publicly owned airports in Canada?
The reverse of this question is asked in
Are there any privately owned large commercial airports?
I arrived at the YWG airport (Winnipeg), checked in, and went in to security.  Already with my items on the conveyor belt, the security personnel tells me that my boarding pass was printed incorrectly (the seat number was wrong), and I have to come out to get it corrected.  While they're working on it, I get told by the police that this airport is privately owned by Winnipeg Airports Authority, and the owner doesn't want me to fly.  It's my first flight through this airport, and I am doing nothing wrong, but the owner of a private property is not required to give a reason; they simply say that if I come back another day, I will be allowed.  There is no way to know in advance whether a given airport will allow me to board, and by now, I already lost roughly $500 and counting because of not being allowed to board.  Since the airline is not involved, there are no laws applying to the situation.  So, no flight rescheduling or compensation is required from the airline.  And if I don't comply, I will be arrested for trespassing.
Is there any way to find out whether an airport is privately owned, so that this does not happen in the future?  At least with a public airport, they would have to reveal their reason if they wanted to kick me out.

If you're interested in trying to be made whole, your story as described above is the sort of thing that local news outlets would love:  "innocent traveler kicked out of airport by capricious corporation".  Might be worth contacting the local CBC, CTV, or Global affiliates.

It sounds like you ended up on a no-fly list, and you probably have some work ahead to challenge it if you want to fly again.  I doubt private vs public ownership is the issue, unless you got really rowdy at this particular airport and got banned from just this one location.  Probably a better question for Legal or Travel SE.

Could this be a case of mistaken identity?

An educated guess would be that the reason is related to the OP's previous experience with Customs and Border Patrol as described in the answer to this question:  https://travel.stackexchange.com/questions/109183/denied-entry-to-canada-will-this-history-deny-me-entry-to-other-countries/125610#125610

Privately owned airports are very much the exception than the rule in Canada.  For starters,

All airports in the National Airports system, with the exception of the three territorial capitals, are owned by Transport Canada and leased to the local authorities operating them. The three territorial airports are owned and, with the exception of Iqaluit Airport, are operated by their respective territorial governments. Iqaluit is operated by Nunavut Airport Services, a subsidiary of Winnipeg Airport Services Corporation, which in turn is a subsidiary of Winnipeg Airports Authority.

In other words, all of the following airports (which handle the great majority of Canadian passenger traffic) are publicly owned:

AB:  Calgary, Edmonton
BC:  Kelowna, Prince George, Vancouver, Victoria
MB:  Winnipeg
NB:  Fredericton, Moncton, Saint John
NL:  Gander, St. John's
NT:  Yellowknife
NS:  Halifax
NU:  Iqaluit
ON:  London, Ottawa, Thunder Bay, Toronto-Pearson
PE:  Charlottetown
QC:  Montréal-Trudeau, Montréal-Mirabel, Québec City
SK:  Regina, Saskatoon
YT:  Whitehorse

Moreover, Transport Canada also owns several small airports in BC, Manitoba, Québec, and Newfoundland/Labrador.
Note that these are only the airports owned by the federal government;  airports not appearing on the above lists may still be publicly owned.  Out of the 20 busiest airports in Canada, there are three that are not on the above list, and none them are privately owned either:

Abbotsford Airport is owned by the City of Abbotsford.
Waterloo Airport is owned by the Regional Municipality of Waterloo.
Toronto-Bishop is owned by PortsToronto.  (This is not exactly public ownership, but it's not private ownership either.)

Sorry, should've mentioned this in the question.  This occurred in the MB: Winnipeg airport.  I was told that the private entity owner is called Winnipeg Airports Authority, and the head of this authority can make such decisions.  Already updated the question.

@Alex:  From what I can tell, the Winnipeg Airport Services Corp. operates the airport under a lease agreement with their parent corporation, the Winnipeg Airports Authority.  They don't actually own it.  You might want to update the title of your question to reflect this distinction.

I do wish to update the question, but unfortunately if I do so the answers would be answering the original question, not the updated one.  So I think I'll leave it as is, and take the answers into account when making future flight plans.

@Alex: you could always ask a new question, or provide an answer to this one that provides more elucidation about what happened if  that would be more appropriate.  If you do ask a new question, best practice would be to link to this one for context.


text:
What is a "pickle switch" in civil aviation?
In an episode of "Air Crash Investigation" the term "pickle switch" was used. I tried to search this term in all the resources available to me and I didn't find any citation, except in some rare cases regarding military aviation.
In any case, it looks like a not standardized term, and I'd like to know if it really exists (and is thus not fictional), why it is named like this, and is it some kind of slang?

Related, nearly duplicate: https://aviation.stackexchange.com/q/82832/54557

The term pickle switch is now generically used for a kind of handheld electronic switch.  They were originally used as the bomb release switch in WWII.
The handle of the switch is a cylindrical object sized to fit in your hand making a fist.  It is sometimes ribbed and generally looks like a dill pickle.  There is a momentary electric switch on the end that you depress with your thumb.
Here is a photo of a Lancaster bombardier holding a pickle switch.  this one does not actually look like a pickle -- I haven't been able to find a photo of one that does.  However, it illustrates the use of the handheld bomb release switch.

Here is an illustration of the bombardier control panel from a US bomber training document.  The pickle switch is stowed on the side of the panel.

In an aircraft, switches mounted rigidly to the airframe can be a challenge because as the aircraft maneuvers, hits turbulence, gets shot, and generally shakes around it is difficult to reach a particular switch.  Furthermore, when the bombardier was heads-down focused on the bomb sight, they needed a bomb release switch they could reliably press with split-second accuracy without diverting attention from the bomb sight.  A handheld switch fits the bill perfectly as the operator knows where it is without having to search for it -- and it is depressed with a simple press of the thumb.
Other applications where an operator needs to be able to press a switch without diverting attention to the switch have come to use pickle switches.  On modern military aircraft, thumb operated switches on the control stick are likely called pickle switches.
I was watching Jeopardy tonight and recognized that the signaling device they use would be considered a pickle switch.

Similar switches are also used by the deck crew on a carrier, so it's not just pilots but also their support crew when performing landing operations.

Agreed, any application where diverting your attention to press a switch would be a problem is likely to use a pickle switch.

A camera cable release is similar too, especially the old mechanical ones, but I've made an electronic one using a 35mm film can and momentary button, to imitate some commercial versions

I always thought they were pickle switches because you used them to "drop the pickles".

In the US, the push-to-talk assembly that air traffic controllers hold in their hand is called the "pickle switch." It is indeed about the size of a small pickle, although it is not cylindrical nor is it ribbed. But in keeping with Rob McDonald's answer, it can be held in a closed hand. Some people have it clipped to their belt, their pocket, or their lanyard, however.

I spent many years in ATC and never heard the transmit switch called a pickle switch.

Perhaps that terminology is more common in Terminal facilities than Enroute?

We used those same headsets in Mission Operations at NASA JSC, but did not use that terminology.

It normally means a switch used to "pickle", or jettison, or release, some sort of load or stores, like bombs or drop tanks (it started as a slang term used by bombardiers).
In the civilian world, you most commonly see pickle switches on helicopters with sling load hooks underneath, for releasing the load when dropping it off, or in flight if an emergency comes up.  Usually a button, or add-on guarded toggle switch, on the control stick, operated by one of the lower fingers.
When flying with a sling load, you need the ability to pickle the load immediately if it starts to oscillate and jeopardize control of the machine. It's pretty unusual to see a pickle switch on a civilian fixed wing airplane, unless it's equipped to drop stores, like a firefighting airplane.

In Boeing aircraft, the pickle switches are the up/down switches on the yoke that operate the trim motor. The official Boeing name is "Stabilizer Trim Switches". "Pickle switch" is the colloquial name.
There are two and both have to be moved at the same time in the same direction in order to make the stab trim move. They are spring loaded to the center (off) position. These are the switches referred to in the Air Crash Investigation series.
This term may also apply to other aircraft.
Anecdote: the yoke mounted stab trim switches on the earlier Boeings operated the trim motor at the highest speed regardless of the speed and configuration of the airplane. There were pedestal mounted switches to operate the trim motor at the lower speed which was desirable in the cruise at high aircraft speeds to avoid abrupt pitch changes. The use of the "pickle switch" during the cruise was frowned upon for this reason and was indicative of an inexperienced or ham fisted pilot.

One prominent accident in which "pickle switches" play a part is Alaska Airlines Flight 261.  In this case, the "pickle switches" are the switches on the yoke that are used to set the longitudinal trim - officially, the "Control Wheel Trim Switches".  See the official report, page 4, footnote 18.

This doesn't really say what it is, though, and thus doesn't really answer the question.

To answer your question, in this case, in civilian aircraft, "pickle switch/button" is absolutely slang and in no way official and as others have already answered, in this case its referring to the pitch trim on the yoke.
I've been flying since 1974, USAF fighters, civilian biz jets and everything in-between and I've never heard it used in reference to the pitch trim, just the ordinance release button ( on the control stick in my case ) of a military fighter/bomber.
Incidentally, in the case of a "runway" trim as happened in the aforementioned  https://en.wikipedia.org/wiki/Alaska_Airlines_Flight_261 incident, you can override a runaway pitch trim with the trim switch to bring it back into trim then pull the circuit breaker to stop it there so it won't go "runaway" again, then, in many, if not all airliners, you have a manual pitch trim wheel in the cockpit to make further adjustments or just manually hold appropriate control pressures from there on. Much better than leaving it in a full deflection position with 120 lbs of pressure required to overcome it.


text:
Why did Avianca have A318s with PW6000 engine?
I want to know about the upgraded PW6000 Avianca A318s. Why did those get upgraded? Did they not produce enough thrust already?

The PW6000 was a second option for engines on the A318, not an upgrade. The rest of the A320 series at the time had options of CFM or IAE engines. The thrust was comparable to the CFM option, but the initial design had worse fuel efficiency than expected, leading some customers to switch their orders to CFM. In the end, only LAN bought A318s with the PW6000 engines, taking delivery of 15. These were later sold to Avianca Brasil, who no longer flies them.


text:
Why does JBU1825 from Ft. Lauderdale to Lima show a large zig-zag in the route?
Why oh why is this non-stop flight flying such a zig-zag? (screenshot from FlightAware)

It isn't, as a screenshot taken later in the flight will show:

The dashed line is FlightAware's attempt to plot the planned route of flight. It's not always accurate, and FlightAware sometimes includes strange zig-zags. If you go to skyvector.com, choose "Flight Plan," and paste in the route shown by FlightAware (BEECH5 BAHMA URSUS UL780 TBG UL780 VAKUD UL780 TRU UV1 ATATU ATATU2), you'll get the more logical route that appears to match the actual flown route:

The skyvector site, along with the flight plan info from flightware, is a very useful combination of tools. Thank so much, Zach.

So this is basically just another example of "you get what you pay for"? Maybe the question should be clarified if the screenshots are from FlightAware, to indicate that fact?

Makes me wonder where in the heck FlightAware gets crazy routings like that. It routes it to the most desolate, uninhabited place in the middle of the Amazon rain forest. There are absolutely no airports, navaids, airways or waypoints anywhere near there. Maybe a few hidden landing strips for drug runners.

Per skyvector it looks like there's another intersection called ATATU over there, evidently FA is getting confused between that one and the one in Peru.

(ignoring FlightAware issues)
Possible reasons for such a deviation:

When air traffic controllers in a country are on a strike then airliners have to fly around that country.
When there is weather en route then pilots need to deviate around that which can be a substantial detour


text:
Why are passengers requested to close their window blinds during a night-time takeoff?
Why are passengers requested to close their window blinds during a night-time takeoff?
I understand why they are asked to open them during day-time take-offs and landings - probably because rescue teams can look inside in the case of a crash. But what´s the point in closing them at night?
The only reason I can think of is that people might mistake the cabin lights for landing lights, hence misjudging the planes orientation. Still, it is very difficult to mistake a row of cabin lights for landing lights?

this questions stands on the assumption that closing the blinds during nighttime is mandatory. where is this claim made?

As I pilot nothing above UL, I can´t cite the regulation. I can just report from my experience of flights in Europe and Asia, where the Stewardess were quite serious about the issue.

On every nighttime flight I have been on, the window blinds have actually had to be open, and on most flights the lights dimmed. This makes sense because in an emergency evacuation at night you want you eyes to be already adjusted to the darkness. Also, having the window blinds open means you can see if it is safe to open the exit door. I have never heard of the blinds being closed for takeoff.

@Ben exactly my point, as also said in another question around here: http://aviation.stackexchange.com/questions/1536 Never heard of blinds being requested to be shut.

@ZsoltSzilagy I have taken many flights in Europe and Asia and have never seen the blinds closed.  Always open.  Please can you quote an airline>

I second this, never closed, always open! For Safety!

Well, I experienced what the OP had. On flights starting around 9-10pm from USA going to Asia, passengers are told to close the blinds. The reason I can think of is that most of the people will sleep (as it's their night time), but the airplane will enter in day light shortly (in just 4-5 hours as opposed to people's 8-9 hours' night). When this happens, very bright sunlight will start entering the airplane, but most passengers are still sleeping. So keeping the blinds down makes sense.

@farhan, was that during take off or once the plane was in the air?

@FreeMan In my experiences, it was well after the takeoff. After takeoff, flight attendants gave us something to drink, then dinner and desserts. Then they sang lullabies and gently tucked us in. Oh wait a minute, maybe the last part only happened in my imagination.

@Farhan the question did specifically address take-off. Perhaps you were too far off in your daydream about the cute stew to notice?  ;)

@ZsoltSzilagy at which airport(s) and with which airline(s) did this occur?

This is a personal experience rather than an authoritative answer, but I thought it would be relevant.
I fly a few airlines fairly frequently and have done for many years. I've always thought that blinds should be open for safety reasons. However, a few years back I was on a flight with an airline that I'd flown many times before and had always enforced the blinds-open policy. To my surprise everyone was asked to close the window blinds. Within a year the trend had increased to point where all flights by this airline were blinds-closed. Then, a year or so after that, the policy was reverted to blinds-open and has remained since.
My theory is that someone in the senior crew, or crew training, mixed up the policy and started enforcing the wrong one (i.e. blinds-closed). I think this is plausible as the reasons for having blinds-open are not entirely obvious. It would also explain the gradual adoption of the new policy. I guess someone eventually realised the mistake, hence the sudden reversal.
Sorry for the long post and lack of sources. Will delete if deemed not helpful.

I've flown a lot over the years with several different airlines and I've had once or twice years ago where they required the blinds to be closed. Other than those few times I've never encountered this.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How can we calculate the fuel that we need?
Let's assume that we are crossing the Atlantic (about 3,878 miles) with a 737-8 MAX. How can we calculate the fuel that we will use during our flight?

Exact same way you would do it for a road trip in your car; it's just a time, distance, burn rate problem on a larger scale.

Related: How can I calculate the fuel consumption of an Airbus 320-200 at various loads?

You need to know the fuel burn per distance covered which depends, among other things, on aircraft weight, speed, altitude and wind (all of which probably change over time), and then you add it up over the trip distance for the planned route (which is likely to differ from shortest possible great circle route for various reasons like airspace organisation and weather). As mentioned by OverD, regulations require some extra bits on top to cover for unforeseen circumstances. As also observed by OverD,  this is also dependent on the individual aircraft - e.g. old engines might need more fuel than new ones, a dent in the fuselage might create additional drag, an inoperative component might require extra fuel or off-optimum altitude etc. etc.
The manufacturer, in your example Boeing, provides the aircraft‘s performance data (e.g. fuel mileage etc.) to the operators.

I agree with Cpt Reynolds however there are more factors that come to play, such as the technical status of the aircraft (some issues require extra fuel uplift), the age of the engines (performance factor).
And a major part is regulation, almost always you require additional fuel to proceed to an alternate airport, plus additional fuel to fly for an extra 30 mins and you require 3-5% of you original fuel as inflight reserve. And not forget the fuel required to taxi the airplane and use the APU.
But if you factor all the above out Cpt Reynolds answer is spot on


text:
Is overall pressure ratio same as compressor pressure ratio for a gas turbine engine?

For a typical gas turbine engine, turbofan in particular, is the compressor pressure ratio same as the overall pressure ratio?
For modelling purposes can we use the same value of overall pressure ratio for compressor pressure ratio too?
If we do so, will we get accurate results for performance?

The overall compression ratio is known as OPR.
Wikipedia states the definition very well:

overall pressure ratio, or overall compression ratio, is the ratio of the stagnation pressure as measured at the front and rear of the compressor of a gas turbine engine

But, should read … as measured at the front of the first compression process and the exit of the last compression process…
The compression ratio of a compression process (where mechanical work is used to increase the pressure) is known by PR, generally using a prefix. Compressor Pressure Ratio, CPR, or Fan Pressure Ratio FPR.
The overall pressure ratio is thus calculated by multiplying the individual ratios.
$OPR= FPR \times IPR \times CPR $

For a typical gas turbine engine, turbofan in particular, is the compressor pressure ratio same as the overall pressure ratio?

No, but the OPR can be the PR in simple gas turbine configurations known as turbojet engines if there is a single compressor. These types of engines are uncommon these days. Today’s engines (turbofan engines) have some sort of a bypass raio to increase the efficiency of the cycle (large bypass for commercial/transport flight, lower for fighter aircraft or business jet engines).
Increasing the OPR is done with increasing turbine inlet temperature (TIT) to allow higher efficiencies while enough fuel can be added to the cycle to get enough power to propel the aircraft (either through the fan, accelerating much air by little, or accelerate few air by a large amount).

For modelling purposes can we use the same value of overall pressure ratio for compressor pressure ratio too?

No, in assessing cycle calculations for gas turbine designs (design exploration studies) to come we generally work with the OPR using constraints on the intermediate and high pressure compressors. We generally choose a FPR and an OPR and calculate the needed HPC pressure ratio.
Dr. N. Cumpsty has done some excellent work in this respect.
Turbofan generations are typically characterized by FPR and bypass ratio, OPR and TIT. In order for your calculations to make sense you need to pick/calculate/use the correct values. Increasing bypass ratio and decreasing FPR cannot be unlimited done without altering the OPR and TIT

Modern turbofan engines typically have several stages of compression, the overall pressure ratio is the ratio of the highest pressure to the intake pressure. But the pressure ratios of each compressor can be a different.
The fan and booster add a bit of compression, followed by an optional intermediate pressure compressor and finally the high pressure compressor.
Because of these multiple shafts and therefor having some parts of the compressors not working on their design point you probably won't get accurate results by simulating a single stage compressor.

The only 3-spool engines I know are the large Rolls Royce engines. All GE and I believe P&W are dual-spool and don't have an intermediate compressor and turbine.

That is true, I mixed up the booster and intermediate compressor there.


text:
How is the compliance time in this AD interpreted?

Within 25 flight hours (FH) after the effective date of this AD, or
  within 100 FH since the last inspection, whichever occurs later, and
  thereafter at intervals not to exceed 100 FH

Do we have the same condition in both states of above? If our A/C inspected since the last inspection in 100 FH over or less because of this state "whichever occurs later", we must comply for 100 FH grace period and 25 F/H is omitted?

Break down the timeline logic.  The key word is "later".
AD released today.
Last inspection may or may not have been done 50 hours ago, which pre-AD was voluntary or part of a non-mandatory maintenance program, or part of some recommended Service Bulletin, that is now being mandated, or maybe it was never done at all.  Assume it was done 50 hours ago.
Which requirement is later?  25 hours from today, or 100 hours since the inspection done 50 hours ago?  100 hours from the last inspection is still 50 hours out from today, so it is later than 25 hours from today, so do it in 50 hours.
Say last inspection was 90 hours ago.  Hundred hours later comes up in 10 hours.  But the AD says whichever is LATER.  25 hours from today is the later than 10 hours from today, so that applies.
Or inspection has never been done before.  25 hours from today.
In no case do you have to do it in less than 25 hours.  It's more or less the same as that other one you posted about landings, but with a bit different wording.


text:
Load factor and how does velocity affect it?

An aircraft initially flying level with a velocity of 130 m/s pulls
into:

a turn of radius 300m at constant altitude (i.e. a horizontal circle) with constant velocity;

a loop (i.e. a vertical circle path) of radius 300m, where the thrust is varied to equal the drag at any point (hence, any change in
velocity is entirely due to gravitational effects).

Determine the load factor for (1.) and (2.) when the aircraft is a the 90, 180- and 270-degrees position within the loop.

I have been trying to understand the question, however, I don't entirely understand what they mean. 90, 180 and 270 don't seem like banking angles and specially for part b, does it mean that at 90 degrees it has travelled 300 m vertically and its velocity will decrease based only on the gravity? Also, does the question assume 0 banking angle?

A loop is a vertical circle, so both are not so different from each other.
First, let's check how feasible the assumptions are. The equation for the load factor $n_z$ in a horizontal circle is the vector sum of gravity and the required centrifugal force for a circle with speed $v$ and radius $R$: $$n_z=\sqrt{1+\frac{v^4}{g^2\cdot R^2}}$$
which comes out as 5.83 with the parameters given in your question. This equals a bank angle of 89° and qualifies this circle as a horizontal loop. Note that the limit load factor for aerobatic GA aircraft is $n_z$ = 6. There are not many aircraft that can sustain this load factor without altitude loss or full afterburner.
The angular positions mean the stations along the circle: 90° is a quarter circle and 360° is the full circle. Of course, if thrust is sufficient, the load factor will be the same throughout the full circle in case 1 of your question.
Part 2 is not much harder to answer: Here we need to subtract the kinetic energy loss sustained in the climb or regained in the descent phase of the loop. The potential energy gained when climbing 300 m is mass times gravitational acceleration times height, so the speed loss is $$\frac{1}{2}\cdot(v_1^2-v_2^2) = g\cdot h$$ with $v_1$ the initial speed of 130 m/s and $v_2$ the final speed after the climb by a height $h$. Using the values in your question again, the final speed for the first 300 m is 104.597 m/s or about 105 m/s and for the second 300 m it is 71.638 m/s. So the answer for the speed is 105 m/s at 90° and 270° and 71.64 m/s at 180°.
Now use those speeds to calculate the load factor at those points. Hint: At the start of the loop it is $n_z$ = 6.74.

b) means straight up into a vertical loop with a 300m radius, with the load factor calculated at the 9 o'clock position as velocity is decreasing and G increasing on the up side, 12 o'clock position as velocity and G reach a minimum at the top inverted,  and 3 o'clock position as velocity and G is increasing on the down side, all due to gravity alone.  Bank angle is not a factor and can be considered "level" flight from that perspective.
a) means the load factor in a 300 m radius level turn, at whatever bank angle is required to achieve that radius at a constant 130 m/s.
The question should end with "loop and turn" instead of just loop and I would assume the 90, 180, 270 business also applies to the equivalent locations in the level turn in the case of a) (and I would presume to be the same number at each location in the level turn).
The question is pretty poorly done, but that's how I would interpret it to make do with what is presented.  Between the confusion over loop and turn and the typos like "id" for "is", the exam writer needs some remedial training.  So if you get the answer wrong, and it's important, I would protest it if that is feasible.

I don't entirely understand what they mean. 90, 180 and 270 don't seem like banking angles

They're not.  They're the position within the loop.  Where the question reads:

Determine the load factor for (1.) and (2.) when the aircraft is a the 90, 180- and 270-degrees position within the loop.

I would suggest parsing it as:

Determine the load factor for

(the turn)  and
(the 90, 180- and 270-degrees positions within the loop).

.

Does it mean that at 90 degrees it has travelled 300 m vertically and its velocity will decrease based only on the gravity?

Yes.  From the entry (at the 0-degree position) to the 90-degree position, it's moved upward by one radius, which is 300m.

Also, does the question assume 0 banking angle?

There will be a non-zero bank angle in the turn.  I would assume a zero bank angle for the loop.


text:
Helicopter total power required: zero wind speed while climbing --vs-- down wind speed while hovering
I've been stumped on this for a while. Let's say you have a helicopter in climb mode (i.e., 100 ft/min) with zero wind speed present. Would the total power requirement be the same as if the helicopter was hovering, but with 100 ft/min wind speed in the downward direction? Note, I'm assuming steady (constant velocity) flight.
Blade element theory (BET) would say that the power requirements would be the same due to equivalent torque on the rotor (i.e., the blade element for both cases will experience the same angle of attack and magnitude of the relative wind). However, in the former case isn't the helicopter gaining gravitational potential energy, so there would be an additional climb power term equivalent to the weight times climb speed? Or does this term get excluded with BET analysis? Can I think of both scenarios as if the helicopter is gaining "air-referenced" potential energy, where it's climbing relative to the air, meaning that the calculated power requirements would be the same regardless?

I suppose that in the first case you have energy associated with a) compensating for the weight, b) winning the aerodynamic drag and c) gaining height. In the second case you have energy associated with a) compensating for the weight, b) winning the aerodynamic drag and c) moving air downward. But I'm not a physicist.

Yes, the power required to hover in a 100 ft/min downdraft is the same as the power required to make a 100 ft/min climb in calm air, all other things being equal. And yes, the helicopter that's climbing is gaining potential energy.
What you're missing is that the downdraft is removing potential energy from the helicopter. If we replace the helicopter with a neutrally-buoyant balloon, it's easy to see that a 100 ft/min downdraft would cause the balloon to descend at 100 ft/min, removing potential energy from it.
To answer your last question, yes, blade element theory completely ignores the effect of gravity. BET is solely concerned with the performance of propeller blades, so, in fact, it ignores the effects of everything besides the blade and things that directly affect the blade.

Thank you for your answer. It makes perfect sense to look at it this way. So with wind present, climb power is more accurately determined by weight times vertical airspeed (not just vertical ground speed), just as parasite power is determined by drag times horizontal airspeed in steady, level flight, correct?

@AdamYassine: Aircrafts move in respect to air, not ground, and therefore everything should be calculated in respect to air. Longitudinally, vertically and laterally. For example in the usual $L=½ \rho V² S C_l$, speed $V$ is measured in respect to wind, not ground.


text:
Boeing 777 PCU Synchronization of Duplicate Hydraulic Actuators for Control Surface Deflection
I have a question regarding the redundancy of hydraulic systems used to deflect the control surfaces of a Boeing 777 aircraft (or the like). We know that "the elevators, ailerons, and flaperons are controlled by two actuators per surface [and] the rudder is controlled by three" [1].
Let's say we have two duplicate hydraulic actuators that control one of the ailerons. How does the Fly-By-Wire (FBW) system ensure that the movement of both actuators are synchronized; in other words, how do the Power Control Units (PCUs) prevent over-extension of one actuator relative to the other so that torsional stresses on the aileron are not exceeded?
I read about LVDT sensors for position feedback being used to control the displacement of the hydraulic pistons with precision. In a hypothetical case we can say that a simple proportional controller would provide a correction to minimize the error, and this error is proportional to the difference between the commanded displacement and the actual displacement of the actuator. Due to differences in resistance between each hydraulic actuator, would a simple proportional control scheme not be sufficient? I envision a more enhanced, "coupled" control scheme to ensure synchronized movement of both actuators:
\begin{equation}
    PCU_1 = K(p_c - p_1) + G(p_2 - p_1)^n
\end{equation}
\begin{equation}
    PCU_2 = K(p_c - p_2) - G(p_2 - p_1)^n
\end{equation}
where $PCU_1$ and $PCU_2$ can represent the voltage signal of the solenoid-controlled servovalve for actuators 1 and 2, respectively. $K$ and $G$ are arbitrary proportional gains. $p_c$ denotes the commanded (synchronized) position of both actuators. $p_1$ and $p_2$ are the actual positions of actuators 1 and 2, as measured by their respective LVDT sensor. Lastly, the exponent $n$ can be an arbitrary integer; the higher the value the more the controller will try to compensate any deviations of $p_1$ and $p_2$, so that the difference $p_2 - p_1$ converges to zero more quickly.
Does this coupled logic sound reasonable? Do current commercial jet transport aircrafts employ a similar PCU control scheme for synchronization of multiple actuators connected in duplicate fashion?
*Note that in the above scenario I am assuming each actuator is operating in Active mode, and the servovalves have a closed center configuration to allow for hydraulic lock when no further Actuator Control Electronics (ACE) command is needed. Also note that each actuator has its own independent hydraulic circuit, pump, reservoir, valve, sensor, wiring, etc. In addition, I'm assuming the actuators are separated along the aileron's span. Each control rod deflects the aileron at a common hinge point and equivalent moment arm, hence why I chose $p_c$ to be the same for both actuators. (The commanded positions of actuators 1 and 2 could still be independent.)
[1]: Boeing B-777: Fly-By-Wire Flight Controls

It would make sense to let each actuator have it's own position sensor simply to make the wiring simpler.

The point of multiple independently-controlled actuators is to provide voting between multiple control systems without a single point of failure. The fact that you have a single command, $p_c$, defeats this. To use this technique, you simply make the structure of the aileron strong enough to withstand conflicting commands. Some systems use self-monitoring: if the surface isn't moving to the position the loop wants, then the loop realizes it's being out-voted and it trips itself offline.

@user3528438 sorry for not making this clear. From the scenario I'm assuming each actuator has its own unique hydraulic circuit, valve, sensor, wiring, etc.

@user71659 when the loop trips itself offline does it immediately switch to Standby mode (i.e., Bypassed or Damped mode, as described in [1])?

Check out section 11.8.5 in the document you linked; that seems to discuss the scenario you're asking about.

@RalphJ WOW... I can't believe the answer I was looking for was right under my nose! Thank you for pointing out that section, which is something I overlooked. It makes a lot more sense to monitor the difference in fluid pressure of the two actuators--as opposed to the difference in their positions--to avoid overstressing the individual control surfaces due to unequal actuation forces.

In other words, it's more efficient for the Primary Flight Computer (PFC) surface control laws to eliminate the so-called actuator force-fight condition via Delta P monitoring, as this variable would be more sensitive to changes in load than the relative displacement of the actuators.


text:
Why are jet engines still so noisy?
In general, jet engines on passenger aircraft make so much noise that in a long-duration flight it becomes really tiresome and uncomfortable. Nowadays, technology is much more advanced than before. Manufacturers have added many features to control the noise.  Then, why do jet engines still produce so much noise?

I am not looking for what is causing the noise as mentioned in the linked question. I want to know why it's like that. Agree that there are things to prevent the noise but the noise is still pretty high. Enough to create a headache.

You may be interested in this article which not only explains what are the causes and what was done, but also what is currently researched to improve the situation, known to be unsatisfactory (not the noise within the aircraft, but the noise for overflown inhabitants - but solving this problem would benefit to passengers too). Introduction: "noise suppression has become one of the most important fields of research due to airport regulations and aircraft noise certification requirements."

Because physics. Understanding the physical causes of the noise is the only way to address this question. If a qualitative explanation is not enough to satisfy you, the next step would be to quantitatively model these causes and see what levels of noise they produce. If you don't want to dig that deep and still feel you don't have a satisfactory answer, then you would just be arguing against physics.

I actually find modern aircraft engines (A380 and newer) to be TOO quiet. You can hear a little kid carrying on 10 rows back - give me a smooth engine roar any day.

Respectfully, "because physics" sounds to me like a demeaning insult to the OP, and it also seems to imply that the problem of noise on an aircraft is an unsolvable problem, similar to "Why can't we travel faster than light?  Because physics."  It might be impossible from preventing a fast-moving jet aircraft from being noisy on the outside "because physics", but the noise on the inside could surely be reduced by adding more insulation, although doing so might not be economical.  But "to add more sound insulation wouldn't be economical" isn't covered by "because physics".

Briefly: inside a jet engine's core or within its fan shroud, huge quantities of air get sucked through while their flow directions get rapidly changed up. This causes parcels of air to bounce off the walls of the engine passages and off their neighbors, a process that generates random (roaring) noise.
In addition, in the center of the engine's core, flammable fuel is continuously injected and vigorously mixed with air, and then set on fire. This too creates a roaring and hissing noise.
The compressor and turbine blades tend to shed sound waves as long as they are spinning around. For slow speeds (like a 2-bladed helicopter rotor, for example) each full rotation of the rotor sends two sound waves your way and you hear a whap-whap-whap noise. For lots of blades spinning at very high speeds, a screaming-siren noise is produced instead.  On top of this, the fan on a high-bypass turbofan will emit a groaning moan when it comes up to full speed.
Then, at the exhaust port of the engine, the bypass, core, and ambient air flows get violently mixed which adds to the hiss and roar of the engine.

Did you mean two sound waves as broadband and rotational?

A large part of the engine noise perceptible outside the engine is not created in the engine, but at the exhaust by the vortexes generated when the two engine flows are released in the free stream. All velocities and temperatures are different, air cannot mix smoothly. This is why some manufacturers pre-mix the 3 flows using serrated ducts. What is heard outside the engine is first the noise of turbulent air.

@mins, thanks, will edit. -NN

See also this article with nice pictures.

A turbine rotor fan spins at 10,000 RPM. With a diameter fan of 123 inches, it has a circumference (Pi x diameter) of 386 inches meaning it spins so fast  on it's outer circumference it goes thru the air at 3,600+ miles per hour (Mach 4.8) , that's a huge amount air displacement and sound comes from air displacement

Mach 58!? Rule of thumb: if you get a result > Mach 10, something is wrong with your calculation or this belongs on Space.SE. I get 3655 mph for your numbers. Converting that to a Mach number requires knowing the temperature (since speed of sound is temperature dependent).

Your calculation went terribly wrong somewhere along the way, I'd blame it on the funny units. When the fan has a diameter 312 cm, it has a circumference of 9.81 m, meaning it spins so fast that the tips move at 1636 m/s, which is approximately Mach 4.77 at sea level conditions (speed of sound 343 m/s).

My mistake i forgot the decimals and forgot to divide 10,000 rpm by minutes.
Pi x Diameter = circumference
π x 123 inches of a GE90 engine fan. circumference is 386.4 inches (32.2 feet) with a revolution of 10,000 rpm (166 per second) to complete a revolution of 32.2 feet  166 times.


text:
What are the advantages and disadvantages of fluidic thrust vectoring on aircraft?

The Rockwell XFV-12 was a research project for combining supersonic speed with Vertical Take-Off and Landing, From wikipedia:

The XFV-12 used a thrust augmented wing concept... in which exhaust would be directed through spaces in a wing opened up like venetian blinds to increase available lift, somewhat like Lockheed's unsuccessful XV-4 Hummingbird.

So I've seen a little bit about the "Rockwell XFV-12 Fighter" plane and "Jetoptera's Bladeless Propulsion System" for a while and I found these types of aircraft and propulsion systems quite intriguing.
However, I couldn't find many aircraft that use this type of propulsion system. Even less information comparing to conventional planes.
How efficient is this type of propulsion compared to conventional systems?

"Thrust augmentation" is generally understood as temporary extra-thrust by using an afterburner or another method. XFV-12 used a thrust augmented wing" which is a wing using venetian blinds to redirect engine gases vertically for VTOL. It's not the same.

This question essentially requires a reader to watch multiple YouTube videos in order to know what's being asked; that's rather time-consuming, and the credibility of videos on YouTube runs the spectrum from extremely credible to not-at-all.  I don't think it makes for a good question to ask for commentary on such videos, especially given how little info is apparently out there on the subject generally (i.e. the OP's 2nd paragraph).  Add to that the comment above by @mins, this comes off as a very low-quality question. DV'd accordingly.

OP means thrust vectoring. Have edited the title and voted to keep open.

Does this answer your question? Why is Russian super-maneuverable thrust vectoring more prevalent than American?

@RalphJ That’s a question with some great answers, it focuses more on supermanoeverability though. The Harrier, and the engineering difficulties that failed the XVF-12, were not mentioned,

@Koyovis then specific details about the XVF-12 angle should be added to the question, not left in a link.

As an aside: in this link several mentions are made of thrust augmentation for the XVF-12.

What is shown in first video is thrust vectoring, the second video seems to be of no practical aeronautical engineering value at all. Thrust vectoring has two major applications:
1. Vertical take-off
Image source including license, cropped photo
Redirecting thrust vertically, for vertical take-off and/or landing. Like the subsonic Harrier does, with four nozzles that can rotate over a range of 98°.

Advantage: tiny runways required, enabling building much smaller and affordable aircraft carriers.
Disadvantage: Four nozzles required, splitting the exhaust from a single engine. Difficult engineering problems as demonstrated by the XFV-12, and piloting problems by having to keep the nose into the wind at vertical take-off. The STOVL F-35B Lightning II uses a separate shaft driven lift fan. Image source and credits

2. Post Stall Technology.
PST is for manoeuvring during combat, as discussed in this question mentioned in a comment by @RalphJ. Supermanoeverability like in the Pugachev Cobra manoeuvre demonstrated by the Su-27 at the 1989 Paris Air Show. An innovative feature at the time, with two camps polarising on the issue at the time according to AIAA paper 91-3194 and expressing the advantage/disadvantage:
 Image source

Pro:

"The key to success in combat with all-aspect missiles is to shoot first. Supermaneuverability allows a pilot to gain a shot opportunity earlier than with conventional maneuverability.

Against:

"Because of the high energy bleed rate, I believe the manoeuvre has no tactical value. It is just an airshow stunt".

Pugachev Cobra - on Youtube

How efficient is this type of propulsion compared to conventional systems?

Thrust vectoring for VTOL implies re-directing the exhaust flow over 90°, and distributing it over at least three support points. And this redirection is associated with considerable losses as experienced in the XFV-12 program, plus stability problems.

The XVF-12 could not produce enough thrust for VTOL and had stability issues. From this link, commentary from people involved in the design:

Just a couple of comments from someone who was involved in the original concept, proposal, and early development of the XFV-12A. The basic problem was with the canard configuration. In order to have the CG in the correct place to maintain positive longitudinal static stability in conventional flight, AND to balance the wing and canard lift forces in hover, the canard augmentors had to produce proportionally quite a bit more lift in much smaller surfaces and ducting than the wing augmentors. In order to do this, the AUGMENTATION RATIO, or the ratio of the augmentor lift force to the lift force of the basic airflow without entrained air, had to be much larger for the canards than for the wings. Couple this with the losses in the canard ducting due to the much longer duct runs, smaller ducts, turns, etc., and that balance was not achievable with any practical useful load.

The drawing below from the wiki shows the arrangement of the four nozzles of the Harrier, the two front ones driven by the bypass fan and the rear ones by the hot exhaust. From the wiki:

The Harrier has been described by pilots as "unforgiving".[61] The aircraft is capable of both forward flight (where it behaves in the manner of a typical fixed-wing aircraft above its stall speed), as well as VTOL and STOL manoeuvres (where the traditional lift and control surfaces are useless) requiring skills and technical knowledge usually associated with helicopters. Most services demand great aptitude and extensive training for Harrier pilots, as well as experience in piloting both types of aircraft. Trainee pilots are often drawn from highly experienced and skilled helicopter pilots.[N 8][12]

It would be interesting to see what else was on display at the 1089 Paris Air Show. :D (some typos are so good, they're not worth correcting!!)

@FreeMan There were stones flying, and a people brought their trained albatrosses.


text:
Do the landing gear modules rotate?
I'm particularly thinking about larger aircraft with more than two sets of main landing gear.

It's pretty obvious you would not want them to rotate while landing and take-off, but while taxiing around or being tugged, the drag caused by fixed axis wheels would be significant on tighter corners, and would leave rubber on the tarmac.
As such one would presumably want the wheels to track the turn.
That effect would be even worse for something like the Antonov 225... which I have no idea how they turn on the ground...

If they do somehow pivot, is there some sort of locking mechanism that is employed prior to take-off and landing.
Note this has nothing to do with crabbing for landing, but for ground maneuvers.

Possible duplicate of Why don't modern aircraft have landing gear that can swivel to compensate for crosswinds?

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